ps1sol

# E x 0 see figure 1 for plot 5 3 22 23 z ie 42

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Unformatted text preview: 1: 2 2 2 2 (16) 2 Figure 1: The region given by |z + 1| &lt; |z − 1| −→ 4x &lt; 0 −→ x &lt; 0 For α = 2: −→ 0 &lt; 3x2 − 10x + 3y 2 + 3 −→ 0 &lt; 3 x2 − −→ In summary, α = 1: α = 2: 4 3 (17) (18) (19) −3· 25 + 3y 2 + 3 9 (20) (21) 10 25 x+ 3 9 2 4 3 2 &lt; x− 5 3 + y2 (z ) &lt; 0, i.e., x &lt; 0 (see Figure 1 for plot) 5 3 (22) (23) &lt; z− , i.e., 42 3 &lt; x− 52 3 + y 2 (see Figure 2 for plot) Note: we can understand these results geometrically. For example, the α = 1 inequality describes the set of z -points such that the distance between −1 and z is less than the distance between 1 and z . The imaginary axis is equidistant from −1 and 1; points to the left of the imaginary axis are closer to −1 than to 1. Problem 3 (20 points) Consider two complex variables z and w. 3 Figure 2: The region given by |z + 1| &lt; 2|z − 1| a) (10 pts) Prove the triangle inequality |z + w | ≤ | z | + |w | using algebraic means. b) (2×5 pts) Use this relationship to obtain the alternate forms: |z | − |w| ≤ |z − w| and |z | − |w| ≤ |z + w| for which you should also provide geometric interpretations. Solution 3 a) Observe that 0 ≤ | (¯w)| z Therefore, (z w) ≤ | (¯w)| ¯ z ≤ Thus, |z + w| = (z + w) (¯ + w) z¯ = z z + ww + z w + z w ¯ ¯¯ ¯ = |z | + |w| + 2 (z w) ¯ ≤ |z | + |w| + 2 |z w| , ¯ ≤ (|z | + |w|) , 4 2 2 2 2 2 2 (24) | (z w)| + | (¯w)| = |z w| ¯ z ¯ 2 2 (25) (26) |z w | = |z | |w | ¯ so |z + w| ≤ |z | + |w| . Another approach involves writing z = x + iy and w = u + iv . Both sides of the triangle...
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## This note was uploaded on 02/22/2010 for the course ACM 95A taught by Professor Nilesa.pierce during the Fall '06 term at Caltech.

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